Control Systems Engineering - Cal Polyfowen/me422/BookUpThroughChap2.pdfControl Systems Engineering A Practical Approach by Frank Owen, PhD, P.E. Mechanical Engineering Department

Control Systems Engineering

A Practical Approach

by Frank Owen, PhD, P.E.

Mechanical Engineering Department

California Polytechnic State University

San Luis Obispo, California

September 2015

© by Frank Owen, September 2015

Table of Contents Preface Chapter 1 – Introduction to control systems Chapter 2 – Laplace transformations Chapter 3 – System modeling Chapter 4 – First‐ and second‐order system response Chapter 5 – Stability Chapter 6 – Steady‐state error Chapter 7 – Root locus Chapter 8 – Frequency response Chapter 9 – Designing and tuning PID controllers Chapter 10 – An introduction to digital control (to be added later) Appendix A – Block diagram algebra

Preface

Preface‐1

Preface

Why this book?

This book has been written for controls students at Cal Poly quite simply to save them money. There are

many, many good controls books available, but they have, in my opinion, three flaws.

1) They are very expensive.

2) They are rather reference books than a basic, first book—what one needs when first approaching

the subject. Thus we have found at Cal Poly that we buy a book for a lot of money and then use only

a small part of it. It is not that the parts that we don’t use don’t have any value. They do. But one

doesn’t need to buy a complete reference book to understand the basics and the essentials of a

topic.

3) They are highly mathematical. Controls is a very mathematical topic, perhaps the most heavily

laden mathematically in mechanical engineering. There are many good engineers in industry that

are not particularly adept at mathematics, who practice engineering with as much intuition and

common sense as mathematical adeptness. A mathematical approach to controls loses sight of this,

leaves many people behind, and does not take advantage of the fact that this topic also makes a lot

of sense, a lot of common sense. Thus the approach taken here is to include what math is necessary

but to appeal to common sense and intuition whenever possible. With today’s modeling tools—

read here Matlab/Simulink—a great deal of the math can be skipped and replaced with model

building, to pose and answer questions that start with “What would happen if we…?”

In addition it has always been my conjecture that what we have developed at Cal Poly in our controls lab

would also be very useful to controls engineers in industry. Our lab, while not unique, is very rare. It

brings controls down to earth and teaches controls engineers how to deal with real systems, how to

model them and then tune the models, and how to set up and tune PID controllers for real systems.

These are the essential skills that a controls engineer must have to operate in industry. In my

experience in academia, these essential skills are not often taught. Controls students have their heads

filled with mathematics, indeed the mathematics of complex numbers, but then they are not given even

a starting notion of how such knowledge is applicable in the real world. This book focuses ever on the

real world of controls in industry. It tries never to lose sight of that goal and tries to avoid the alluring

trap of mathematical elegance and indeed mathematical snobbishness that seems common in the field

of academic controls. So the book has also been written for industrial practitioners of control theory

who need to understand the topic and then bring into play to their advantage.

The other influence that led me to write this book was the three years I spent teaching controls in

Germany, two years at the Munich University of Applied Sciences and one year at the Karlsruhe

University of Applied Sciences. In Germany textbooks are rare. Rather students work from a script, a

collection of the professor’s notes organized and printed for student use. This book is really a script, a

collection of my notes from teaching controls over the past decade. Though writing a book is a lot of

Preface

Preface‐2

work, it’s common practice to have short, directed scripts at low costs for poor students. So I thought,

why don’t we do the same at Cal Poly? We have lots and lots of experience teaching controls, so we

should be able to come up with a good script. Besides, with our involvement in our laboratory, we have

already demonstrated that we can come up with a high‐quality document for teaching the lab portion of

the course.

Thus it is my hope that students will benefit from this practical approach to controls just as they are

assuredly benefiting from saving almost $200 (in 2010). And I hope that this script serves as an example

of what could be done in other courses at Cal Poly if professors would take their hard‐won experience,

collect it, and make it available at low cost to those eager to learn but without a lot of money to buy

expensive reference books. This does not mean that one shouldn’t buy the expensive reference books.

Maybe one needs them in his or her work. But at that stage, one has the means to buy them or one’s

company will buy them when there is a need.

The use of Matlab/Simulink

It is hard nowadays to envision practicing controls engineering without Matlab/Simulink. The

employment of this software in analyzing systems and designing controllers—indeed now in running

real controllers in physical systems—is de rigueur. This text does not include a tutorial in learning

Matlab/Simulink. That’s available online or with the software. It is assumed that the reader has some

knowledge of this software. Problems are posed in the text that directly direct the student to use this

software. Occasionally tips are given in specific applications that illustrate the utility of a particular

Matlab command or Simulink procedure. If the user’s knowledge of this software is not at a level where

these references to it make sense, he or she should explore the software a bit, researching its help

facility for background knowledge. Controls requires knowing about only a tiny bit of Matlab and

Simulink. So the reader is not required to do any extensive foundation‐building in order to be effective

with Matlab/Simulink in his or her study of the subject.

Acknowledgements

Wow, how did a hillbilly guy from a lawyer family in Mississippi ever get to the point that he could sit

down and write a controls book almost directly out of his head? Well, if I told that whole story, that’d

be a book in itself. Lots of hard‐won experience but also lots of help along the way. It’s always been my

contention that when bestowing thanks, we never go far back enough. So I want to at least go back and

thank my high school mathematics and physics teacher, Mac Egger. He didn’t plant the original seed,

but he was there close to the beginning. Then there were lots of twists and turns to get to this point.

Along the way: Glen Masada at the University of Texas at Austin taught me classical controls when I

went and got a mid‐career PhD. Just before that I worked at one of the largest coal‐fired plants in the

United States, American Electric Power’s Gavin plant in Gallipolis, Ohio. A plant engineer there taught

me a lot, Randy Scheidler. He taught me a lot about power plants but also about the level of knowledge

of good plant engineers in the United States. Randy served as a sort of model to use in keeping what

I’ve written practical, of trying never to write anything without showing how it is used. And I must back‐

up further and thank the folks at Trax Corporation in Lynchburg, Virginia for their invitation to come

Preface

Preface‐3

work for seven months with them in 1992 on power plant simulators. I learned lots about steam power

plants and how they’re controlled from this experience at Trax.

Since my arrival at Cal Poly in 1998 I have been involved in controls as often as possible. The course

there was handed off to me by Mike Ianci, Ed Garner, and Ed Baker. They had built a very practical,

hands‐on lab. Though we’ve replaced much of the equipment in it, some is still left from those days,

and much of what was added can be viewed as refinements and improvements of what they

bequeathed to us. Here I have the pleasure of working with very practical, hand‐on people like myself—

John Ridgely, Charles Birdsong, Bill Murray, and Xi Wu. All have contributed one way or another to our

work in making this laboratory more practical and hands‐on. Some have had even to deal with the

consequences of what I regarded as a good idea at the time, that required a lot of work on their parts to

implement, to work the bugs out. Through their efforts we have a top‐notch controls lab. I have seen a

better one nowhere, neither in America nor in Europe. It was their sweat that made this lab the great

teaching tool that it has become.

My two sojourns in Germany were important contributors to this book. I had the pleasure of working

with several accomplished controls engineers there. Two stand out. Manfred Schuster at the Munich

University of Applied Sciences welcomed me into his lab and gave me a very concise script to teach out

of. His script is, in fact, a model for mine. At the Karlsruhe University of Applied Sciences I had the

pleasure of working with Helmut Scherf, a committed controls nerd. Helmut is a rare, rare example of a

practical controls engineer. He has published a great book in German of Simulink models of very

common, practical systems. He has built and is still building practical, low‐cost systems for his controls

lab that serve as useful platforms for turning on the controls light in students’ heads. Many of his

perceptive, cut‐to‐the‐quick methods of thinking about controls topics have been incorporated into this

script.

So that’s the story in brief of how this book came about. I hope that you enjoy it and find it useful.

Frank Owen

San Luis Obispo, California, U.S.A.

June 2012

Chapter1–IntroductiontoControlSystems

1‐1

Chapter 1 – Introduction to Control Systems

Goals

The purpose of this chapter is to give you an overview of the topic of control systems and to introduce

you to the basic concepts that you need to go forward. Presented are

Basic control loop anatomy, the parts and pieces of control loops and how they are configured

Positioners vs. regulators, the two basic types of control loops

A fly‐by‐wire system vs. a cruise control system, iconic examples of the positioner and the

regulator

A beginning discussion of block diagrams

PID controllers, the most commonly used controllers in industry

Examples of control systems used in industry

Control theory is a relatively new field in engineering when compared with core topics, such as statics,

dynamics, thermodynamics, etc. Early examples of control systems were developed actually before the

science was fully understood. For example the fly‐ball governor developed by James Watt to control

overspeed of his steam engine was developed out of necessity, long before the science of controls came

into being. Figure 1.1 shows an example of this controller. The fly‐balls are mounted on a shaft that

turns and is driven by the engine through the pulley shown. As the engine speeds up, the fly‐balls are

flung outward by their centrifugal force. This outward movement pulls the lever arm down, which raises

its other end. This is tied to the steam inlet valve, which closes as the flyball weights move further

outward. So if the engine tries to run away, the inlet steam valve will close, shutting off the fluid driving

the engine.

Figure 1.1 – Flyball governor

Many say that the development of the airplane by the Wright brothers was enabled by their

understanding of controls‐‐that and the development of a light‐weight engine powerful enough to


1‐2

propel their machine into the air. Their development of wing warping enabled them to steer their

airplane, something that had been impossible up to that point. And it is certainly true that much of

control theory grew up with the airplane, as airplanes were developed during the two world wars and

also throughout the 20th century for civilian purposes. As jet engines were developed and airplanes

became bigger, it became ever more problematic to pilot an aircraft with just mechanical connections

between the pilot’s controls in the cockpit and the surfaces elsewhere on the airplane that steer it

through the air. Thus the fly‐by‐wire system was developed, which cut this direct connection between

the cockpit controls and the control surfaces on the airplane. In a fly‐by‐wire system, the movements of

the stick or yoke and the rudder pedals in the cockpit are merely sensed by sensors. Electrical signals

are then sent to actuators driving the appropriate surfaces, and then these move the ailerons, the

elevator, or the rudder to steer the plane according to the control inputs made by the pilot. Of course

the force applied by the actuator on the control surface can be many times what a human could apply

directly. And the force can be applied at an actuator far distant from the pilot. So fly‐by‐wire brings

with it the advantage of force amplification and remote control.

In industry one finds control systems of many types. In a refinery, chemical plant, food processing plant,

or a power generation facility one finds control loops for controlling tank levels, pressures of fluids at

various places in a plant, power output, valve position, pump, fan, or turbine speed. Modern‐day fighter

jets actually are designed to be unstable. This allows them to maneuver quickly. They can only fly

because a control system stabilizes their flight, making corrections at a speed that no pilot could match.

If one of these plane’s control system failed in flight, the plane would be unflyable and would crash.

There has been a tremendous growth of control system use in the modern automobile. There are even

now drive‐by‐wire and brake‐by‐wire systems, where, like in the airplane, the direct mechanical or

hydraulic connection between input devices and what they control has been cut and replaced by wish‐

sensing devices and then transmission of an electrical signal to an actuator to turn the wheels or to

apply the brakes. Like the control of the unstable airplane, skid detection and control take advantage of

an automatic control system’s speed. A driver who loses control of his/her car may be saved by such a

system. It springs automatically into action upon detecting a skid situation and applies the correct

braking forces to rescue the car from the skid…and it does this before the driver is even aware that a

problem exists.

Besides these applications, control theory is useful even for analyzing manually controlled systems. A

human operator is in this case actually playing the role of the controller. A human’s sensing of and

reacting to inputs while manually controlling an industrial system or a vehicle is actually a study in

controls. His or her reaction times, the force feedback or the angular travel of a steering wheel or an

operating lever, such human‐machine issues are within the realm of control theory. Mathematical

models of the human controller have even been developed, so that a dynamic model of a manually

operated system can be completed and studied.

Much of what has been discussed here can be illustrated with the example of a pilot in an airplane.

Take the case of an airplane without a fly‐by‐wire system, with direct connections via cables and pulleys

between the cockpit controls and the control surfaces, as one finds in a small, general‐aviation airplane.

A controls expert might study the effect of a human controller during some flight maneuver or critical


1‐3

situation. This system is purely mechanical with a human controller in the loop. But some small planes

also have autopilots, so a controls engineer had to design a system that would sense flight conditions

and operate the controls without intervention by the pilot. Even more complicated is the case of a

larger airplane with a fly‐by‐wire system. Controls engineers designed the sensing‐and‐reaction link

between the cockpit controls and the corresponding motion of the control surfaces. Even when the

plane is being flown in manual mode, a sophisticated control system is engaged simply to sense the

pilot’s movement of the input control levers and pedals and transmit the commanded motion to

actuators that will bring it about. Now consider the case of a fly‐by‐wire system with an active autopilot.

The control system senses flight conditions—altitude, heading, and speed—and automatically operates

the proper actuators—elevator, ailerons and rudder, and throttle—to maintain desired values. Thus

these four variants of doing the same task—flying an airplane—show increasingly complex examples of

modern control systems.

Basic control system anatomy

Classical control systems are SISO systems, single‐input‐single‐output, as opposed to MIMO systems,

multiple‐input‐multiple‐output, which are more complicated. For a control system the input is the

desired value, and the output is the actual value (See Figure 1.2).

Figure 1.2 – SISO control system

A good example is a cruise control system for an automobile. The user inputs a desired value, say 65

mph. Usually one does not type this in. One drives the car up to this speed manually, then pushes a

button. The speedometer senses the speed, stores this in an on‐board computer, and then it is the job

of the cruise control to keep the automobile at this speed.

Thus, when everything is working as it should, the actual value is equal to the desired value. In German

these variables are known as the Sollwert, the should‐value, and the Istwert, the is‐value. In controls it

is always good to bring things down to earth, because controls can get so theoretical, one quickly loses

sight of what is going on or why one is doing what one is doing. I like to refer to these two values as

“what you want” versus “what you’ve got”. When what you’ve got isn’t what you want, then

something’s wrong. When Sollwert – Istwert ≠ 0, then the control loop is not doing its job, and

something is broken or something needs to be changed to make this difference 0.

Actually this difference has a name, the error. That’s not error in the sense of a mistake. Rather it’s

error in the sense of deviation. In a perfectly functioning control system, the error should be 0, and

what you’ve got should be what you want.

Let’s look inside the control loop, at the anatomy of a control loop. Almost all control loops are the

same. They are all made up of five components arranged always the same. Sometimes it is not easy to


1‐4

recognize these elements in an actual system. But it’s always a good idea to try. This structure is

fundamental to control theory and represents the underlying functions that are needed to make

feedback control work. As you probably have already concluded, the basic structure of a feedback

control system is a loop (see Figure 1.3).

Figure 1.3 – Basic control loop anatomy

The five elements are:

1. the comparator

2. the controller

3. the actuator

4. the “plant”

5. the sensor

Let’s discuss these components one by one. I’ll present them in the order that’s easiest to use to

identify them in a real system. Usually the easiest element to identify is the sensor. For a cruise control

system, the sensor is the speedometer. The sensor always measures the actual value and then feeds it

back to the comparator to compare with the desired value. The comparator is just the summing block

that takes as input the desired value and the measured value. That’s the nature of feedback control,

and that’s why it’s called feedback control: the actual value is fed back to the desired value and

compared. Another common example is the thermostat in your house. A thermometer in your house

measures the interior temperature and then compares that with the desired temperature you have

somehow entered on the faceplate of the thermostat.

The error signal is the output of the comparator. It is also the input to the controller. As you can see, all

of these blocks in the block diagram of Figure 2 are SISO blocks, and each output becomes the input of

another block. The controller takes the input from the comparator, the error, and decides how the

system should respond. If the error is 0, then what you’ve got = what you want, and the system should

do nothing. If the error is not 0, then the controller should take some action.


1‐5

Nowadays, with digital controls, the controller is usually just a piece of software running in a computer

somewhere. For a cruise control, there is a computer algorithm running in an on‐board computer that

performs this task. So if someone asked you to point to the controller in the cruise control loop, you’d

have a hard time doing that without talking with the engineers that designed it. Or you could just point

at a black box in the car and say, “There it is”, and most people would have a hard time disputing this.

If the error is not 0, then the controller needs to take action. Eventually it wants to influence the plant.

This is a funny term for the thing that we actually want to control. But control theory grew up in

industrial plants, so that is why this block has this name. The plant can be hard to identify. One

identifies it often by asking, “What are we trying to control?” and then the plant is the thing that that

value is a property of. For example in a cruise control loop, the speed is what we are trying to control.

And the speed is a property of the car. So the car is the plant in a cruise control loop.

Often the actuator is the hardest component to identify, so often we leave it for last. Often it is hard to

draw a line between the actuator and the plant. Often it’s hard to answer the question, “Where does

the actuator end and the plant begin?” You’ll see this dilemma with experience. So good questions to

ask are “What does the controller talk to?” or “Where does the controller send its signal?” or “By what

means does the controller influence the plant?”. In a cruise control system the plant is the car. The

actuator is the throttle. Or maybe it’s the engine. It’s the thing that causes the car’s speed to increase

when the controller notes that the car is going too slow and needs to speed up. What you have is less

than what you want, so do something.

Thus these five logical components are always present in a classical control loop. Though it may be

hard, it is always of value to try to identify the physical components that correspond to the logical

components.

Note also in Figure 2 that the signals between the blocks also have names:

1. r – desired value (“r” stands for reference value; this is also known as the controller setpoint)

2. e – error

3. u – command

4. f – force

5. c – actual value (“c” stands for controlled value)

6. b – measured value

These variable names are not standard by any means, but one sees them often. You should be aware

that often variations of them are used. But we need names for them so that we can refer to them when

talking about what’s going on in the loop.

Two types of control loops: positioner and regulator

Control loops come in two flavors—positioners (also known a trackers) and regulators. Both are made

up of the same components presented above. What differentiates them is actually how they are used,

what their purpose is. The control loop shown in Figure 1.3 is a positioner. This loop can be modified to


1‐6

configure a regulator, as shown in Figure 1.4. Note that the difference is the addition of the disturbance

between the actuator and the plant. This hints at the difference between the two loops. A positioner

has a desired value that changes often. A user is operating a plant through the control system. It is the

job of the control system to sense the operator’s wishes and drive the plant to the point that the user

desires. In contrast, in a regulator system, the user wants normally that the actual value stay at some

preselected level, even though external influences are working to drive the system off of the preselected

level. A good example is the cruise control system of a car. You set the desired speed to a fixed value.

But upward and downward grades tend to make the speed deviate from its desired value. It is the job of

the control system to keep the system at a preselected speed in the face of disturbances that tend to

deflect the actual speed from this wished‐for speed.

Figure 1.4 – Regulator loop

It is common to arrange control loops so that the input is on the left‐hand side and the output is on the

right‐hand side. With a regulator loop, when the desired operating level is chosen, the loop selects this

level as the reference level and considers it to be 0. The loop works in deviations from this operating

level. We shall see later how this is done. At present it suffices to note that the desired value is the 0

reference value, so the r input can be eliminated. The loop can be reconfigured as shown in Figure 1.5.

Here the input is the disturbance, and the output is still the same variable of interest, c. The reworking

of loops as shown in this example is commonly done in controls and is known as block‐diagram algebra.

We shall see many more examples of this in the material to come.


1‐7

Figure 1.5 – Reworked regulator loop

PID controllers, the workhorse of the industry

PID (Proportional‐Integral‐Derivative) controllers are by far the most common controllers used in

industry. The name refers to three different actions that the controller makes in responding to a non‐

zero input, the error, as we have seen above. Thus we speak of proportional action, integral action, and

derivative action. The three actions occur simultaneously. The configuration of the controller is a

parallel configuration, as is demonstrated in Figure 1.6.

Figure 1.6 – PID controller configuration

Note in the figure that the input signal, the error, is first treated one way or another and then multiplied

by a constant. The top path is the proportional path. Here the output is proportional to the error,

hence the name. There is no action taken on the input signal. It is just multiplied by KP and then passed

on downstream to the output. The integral action is the second path. Note that the error (the input) is

first integrated. The output of the integrator block is the integral of the error. Thus if you plotted the

error curve vs. time, this signal would represent the net area under this error curve through time. This is

then multiplied by KI and becomes the integral action. The derivative action is the third path. Note that

the error is first differentiated. The output of the derivative block is then not the error but the rate of

change of the error at the current time. This change rate is then multiplied by KD to become the

derivative action. All three actions are added together in the summing block to become the total PID

controller action.


1‐8

Why one would do this is at this point not clear at all. But as we shall see, each of these actions has a

specific use or justification and usually improves the control response. The three constants—KP, KI, and

KD—are called the controller gains. KP is the proportional gain, KI is the integral gain, and KD is the

derivative gain. It is also often the case that one of the actions is not present. As we shall see, the

proportional action is by far the most sensible and useful action. Often controllers have only P action—

that is KI and KD = 0. We call these P‐only or just P controllers. A controller with no D action is called a PI

controller. One with no I action is a PD controller. So we encounter P, PI, PD, and PID controllers. Note

that all of these have P action. There may be an oddball case without P action, but that is what it is, an

oddball case.

Problems

1.1 Make a conceptual model of a brake‐by‐wire system. The force on the brake pedal is sensed as a

desired braking force. The greater the pedal force, the greater the force applied by the brake pads

to the brake discs. This measured pedal force is sensed by a load cell, which produces a voltage

proportional to this force. This voltage is delivered to an interface board, which converts it into a

digital number in a microprocessor. The controller running in the microprocessor produces an

output signal that is then converted into a voltage that drives an electromechanical actuator. This

drives a piston, the master cylinder, and produces a pressure. The pressure works on the brake

piston that applies the braking force to the disc pads. The braking force is measured using actually

a pressure sensor. Knowing the size of the brake pads, the braking force can be determined. The

force applied to the brake pads is not necessarily the same force applied to the pedal, but they are

proportionally related. Make a block diagram of this system, showing how all the components fit

together to compose the system. Each block should all contain the name of a system component.

Each line between the blocks should show the type of signal being transmitted between blocks.

1.2 The figure below shows a part of the electrical power generation system in a conventional steam

power plant. Fuel (gas, oil, or coal) is supplied on the left‐hand side to the boiler. There water is

heated into steam and stored in a steam drum. From there the steam flows through a control

valve into the turbine, which turns the plant’s generator. The generator produces electric power.

There are several SISO loops in play here. One loads the generator by increasing or decreasing its

electric field. Of course when more electricity is needed, more fuel will be needed to support this.

But the connection between fuel in and electricity out is not direct. The three “lollipops” shown

represent measured quantities. Think about what these might be. Then write in the blocks the

quantities that are measured. Draw dashed lines from these lollipops to the actuators that control

them. It is a chain of events that lead from more power required to more fuel supplied. Consider

the case of a desired increase in power out. Write out in words the sequence of cause‐and‐effect

events that will lead the steam plant to a new, higher level of operation. Make a copy of the

completed diagram to complete the deliverable for this problem.


1‐9

1.3 Sometimes a plant is a two‐part plant, and a disturbance enters the plant midway between these

two parts. Draw a regulator loop for such a plant with a negative disturbance entering between

Plant 1 and plant 2. Let the disturbance be the input to the loop.

1.4 Create a Simulink model of a PID controller. For this you will need to use gain blocks for the three

controller gains. Use an integrator block and a differentiator block for the integral action and the

derivative action. At first just set all gains to a value of 1. As input, use a constant block and set

its value to 0. All three control actions are summed with a sum block. Use a scope block at the

output of the sum block to capture what output the controller delivers over time. Also place

scope blocks on all three control actions to see how they behave. Of course in the real world the

controller would be hooked into a system and receive the system error as its input. Its output

would be fed downstream to an actuator. With 0 as a steady input, you have modeled the case of

a control loop where the actual value is equal to the desired value. What should the controller do

and what does it do? Now make the input 1. What does the controller do now?

Chapter 2 – Laplace Transforms

2-1


Goals

To describe what Laplace transforms are

To show one how to solve ODEs using Laplace transforms

To show the necessity of using inverse Laplace transforms

To show the method of partial fraction expansion for finding the inverse Laplace transform

To explain what a transfer function is

To explain the Final Value Theorem as a method of finding a system’s steady-state response

The rôle of Laplace transforms in the age of Matlab/Simulink

Control systems are used to control the behavior of mechanical, fluid, electrical, and other types of

physical systems. These systems exhibit specific behavior over time that follows the dictates of physical

laws, such as Newton’s Second Law, Kirchhoff’s Laws, flow continuity, etc. The application of such

physical laws to the system at hand results usually in a linear ordinary differential equation as an

expression of the behavior of the system over time. The solution to the ODE is the time history of the

variable of interest. For a very common example found in motion control systems, a mass-damper-

spring system is subjected to an actuator force to drive it into a new location. Such a system is shown in

Figure 2.1.

Figure 2.1 – mck-system

If one analyzes this system using Newton’s Second Law, one arrives at the system’s equation of motion,

which is a linear ODE in x and its derivatives.

��̈ + ��̇ + �� = �(�)

The solution to this equation is x(t), the time history of the motion of the system under the influence of

the applied force. This is a typical example found often in a motion control system, and this is an

important sub-branch of controls. In automated manufacturing systems and in aircraft control systems,

for example, the aim of a control system is to move a mass from one position to another. The topic of

the next chapter is to cover mechanical systems such as this as well as systems in other domains

(electrical, fluid, flow). The analysis of all these systems using physical laws results in corresponding

ODEs. Thus the solution of these ODEs is pertinent to getting the time behavior of these systems.

Prior to the advent of such simulation systems as Matlab/Simulink, the main solution procedure for

getting these time histories from the ODE was Laplace transforms. Much of the mathematical drudgery

of this approach can be avoided nowadays, with the use of Matlab/Simulink. So the necessity of an in-

depth coverage of Laplace transforms has been obviated. Still, anyone involved in the science of control

theory needs to know the basics of Laplace transforms, because their use explains much about the


2-2

behavior of an ODE-modeled system. Thus this chapter aims to introduce you to Laplace transforms but

not overburden you with details that you will not need in this day and age of powerful, inexpensive

simulation tools.

The Laplace transform and its use

One writes the Laplace transform of a function as shown.

�(�) = ℒ{�(�)}

The actual mathematical definition of the Laplace transform is

�(�) = ℒ{�(�)} = � ��(�)��

�

But as you shall see, for all common functions F(s) has already been worked out, so there is hardly ever

any need to resort to this definition. Thus one uses tables of Laplace transforms to solve ODEs. A table

of Laplace transforms is given in Table 2.1. The unit impulse, step, and ramp functions are common

input functions for control systems, as we shall see.

f(t) F(s)

1 1

�

(t) (unit impulse function) 1

u(t) (unit step function) 1

�

t·u(t) (unit ramp function) 1

��

�� ∙�(�) 1

� + �

sin(t) ∙ u(t) �

�� + ��

cos(t) ∙ u(t) �

�� + ��

�� ∙sin (��) �

(� + �)� + ��

�� ∙cos (��) � + �

(� + �)� + ��

��

(� − 1)!� ∙��

1

(� + �)�, � = 1,2,3,…

Table 2.1 – Laplace transforms


2-3

Thus we see that functions are transformed from the t-domain into the s-domain by the Laplace

transform. This offers advantages and a means to solve differential equations. Also useful is to see how

operations such as integration and differentiation transform into the s-domain. Table 2.2 gives Laplace

transform theorems for important operations.

t-domain s-domain

a·f(t) (linearity) a·F(s)

f1(t) + f2(t) (linearity) F1(s) + F2(s)

��

�� ∙�(�) − �(0)

��

�� ∙�(�) − � ∙�(0) − �̇(0)

��

�� ∙�(�) − �� ∙�(0) − ��(0) − ⋯− ��(0)

��

��= �� ∙�� ∙�(�) =

1

�∙�(�)

�� ∙�(�) �(� + �)

�(∞) (Final Value Theorem) lim�→�

� ∙�(�)

Table 2.2 – Laplace transforms of operations

The Laplace transform is linear. The implication of this is seen in the first two table entries. How

differentiation translates to the s-domain is seen in the next three table entries. There is a pattern

there. This is given in the entry for the nth derivative. Notice in all these entries that the initial

conditions of the system play an important rôle. Notice also that the pattern of powers of s decrease in

the sequence of the sum, while the order of the derivative of the initial conditions increases in the

sequence. This is very easy to mix up. It is often the case that the initial conditions of a system are all

0—that is, the system is at rest until t = 0. This greatly simplifies the solution, when this is the case. In

this case we can say that differentiation in the t-domain is equivalent to multiplication by s in the s-

domain. Thus differentiation becomes an algebraic operation in the s-domain. This summarizes the

strength of the Laplace transform. Also note what happens when the degree of differentiation is -1.

This corresponds with integration in the time domain.

Solving ODEs via the Laplace transform

As we shall see in the next chapter, one often encounters 1st- and 2nd-order systems when modeling

physical components. The mck-system shown above is an example of a second order system. A 1st-

order system has the form

�̇ + � ∙� = �(�)


2-4

Note how the ODE is arranged. x and its derivatives are on the left-hand side of the ODE. All non-x

terms are on the right. As given, there is a generic forcing function on the right. To find the time history

of x, one must know what F(t) actually is. A common loading function in practice is the step function.

This function represents a sudden, constant load applied to the previously unloaded system. One uses

the linearity of the Laplace transform and the unit step function to model a load of any size. Thus if the

applied load is as shown in Figure 2.2

Figure 2.2 – Step load

its mathematical representation is

�(�) = �� ∙�(�)

So the equation of motion of the system becomes

�̇ + � ∙� = �� ∙�(�)

Let’s see how one would go about solving such an ODE using Laplace transforms. First take the less

complicated case of the system at rest (0 initial conditions) when the load is applied. One uses the

linearity properties of the sum and multiplication by a constant. In the s-domain this equation becomes

� ∙�(�) + � ∙�(�) = �� ∙1

�

We solve this equation for X(s). This is not what we want for a result. We are seeking x(t), the solution

of the ODE, the system’s time history of motion. But solving for X(s) is a step in the right direction. So

�(�) =��

�(� + �)

To get x(t), one looks in the Laplace transform table for an entry in the right hand column that

corresponds to the right side of the equation above. There is none. The problem is that the solution for

X(s) above is a combination of entries. To separate the components above, one uses a process called

partial fraction expansion. The procedure is as is illustrated below.

�(�) =��

�(� + �)=�

�+

�

� + �


2-5

This separation of components in the denominator introduces the unknown constants A and B. To find

them, one first multiplies the equation by s.

�� + �

= � +� ∙�

� + �

Then one lets s→0, which eliminates the second term on the right-hand side from the equation. Thus

� =��

Now multiply the equation by s+a.

��=�(� + �)

�+ �

Now let s→-a. This eliminates the first term on the right-hand side of the equation. Thus

� = −��

So,

�(�) =�

�+

�

� + �=�� ⁄

�−�� ⁄

� + �

One finds the two entries, �

� and

�

�� in the table. Multiplication by the constants

�� and

−�� can be

dealt with with the linearity property. Thus

�(�) = ℒ��{�(�)} =��∙�(�) −

��∙�� ∙�(�) =

��∙�(�) ∙[1 − ��]

This case occurs so frequently, it is called the step response of the system.

Also very common is subjecting a second order system to a step input. This would appear

�̈ + � ∙�̇ + � ∙� = �� ∙�(�)

Again, consider a system that starts from rest, so �(0)̇ and �(0)are both 0. Taking the Laplace

transform of the equation

�� ∙�(�) + � ∙� ∙�(�) + � ∙�(�) = �� ∙1

�

�(�) =��

�(�� + � ∙� + �)


2-6

X(s) seems to be made of two components because of the two factors in the denominator—s and

s2+a∙s+b. Actually the inverse transform back into the t-domain depends upon the nature of s2+a∙s+b,

specifically whether its roots are real or complex. From the quadratic equation

� =−� ± √�� − 4�

2

Case 1 – Real, distinct roots

In this case,

� =��√��

�= −� and � =

��√��

�= −�

Thus

�� + � ∙� + � = (� + �)(� + �)

so

�(�) =��

�(� + �)(� + �)

and one proceeds just as illustrated with the first-order system above. This is an overdamped second-

order system and can be regarded as simply a 2x(1st-order system).

Case 2 – Real, non-distinct roots

If a2 = 4b, the system has 2 repeated real roots. In this case, the previously illustrated partial fraction

expansion must be modified as follows. s = -a/2. So

�(�) =��

�(� + �/2)�

To deal with the double root, one uses the following partial fraction expansion.

��(� + �/2)�

=�

�+

�

� + �/2+

�

(� + �/2)�

One finds A as previously done with the first-order system. Then the procedure differs to find B and C.

Multiply the equation by (� + �/2)� and let s→-a/2.

��= (� + �/2)� ∙

�

�+ (� + �/2) ∙� + �

� = −2 ∙��


2-7

To get B, differentiate the F0/s equation above with respect to s.

−��

= 2 ∙(� + �/2) ∙�

�− (� + �/2)� ∙

�

��+ �

Now B is isolated. One lets s→-a/2 again to leave B alone.

So the trick to dealing with repeated roots is to multiply the equation for X(s) by the repeated-root

factor, then differentiate.

The case of repeated roots is known as the critically damped second order.

Case 3 – Complex roots

If the roots of the quadratic factor are complex, the partial-fraction-expansion procedure is as follows.

�(�) =��

�(�� + � ∙� + �)=�

�+

� ∙� + �

�� + � ∙� + �

To find the unknown coefficients A, B and C, one multiplies through by the denominator �(�� + � ∙� +

�).

�� = (�� + � ∙� + �) ∙� + � ∙�� + � ∙�

Thus

�� = (� + �)�� + �(� + �)� + � ∙�

Recall that a and b are known parameters. So matching coefficients on either side of this equation,

� + � = 0, � = −�

� ∙� + � = 0, � = −� ∙�

� ∙� = ��, � =��

�

� = −� ∙��

� = −��

This gives then

�(�) =�� ⁄

�−

(�� ⁄ )∙��∙(�� ⁄ )

��∙��=

��

��

�−

��

��∙�� (Eq. 2.1)

The fraction in parentheses can be reworked to give


2-8

��

��∙��=

��

��∙��

��

��

�

=�� ⁄ �� ⁄

��

��

��

�

=��/�

��

��

��

�

+�

�

�

��

��

��

�

(Eq. 2.2)

If we compare the two terms of this equation with the last two entries of Table 2.1 and use the next-to-

the-last entry in Table 2.2, we see that the former term represents a decaying cosine and the latter

represents a decaying sine. The results of Eq. 2.2 need then to be combined with Eq. 2.1 to get an

expression, which can be inverted back into the t-domain. The decaying sine and cosine can be

combined using trigonometric identities to yield

�(�) = �� ∙�1 − ��∙��∙� ∙��(�� ∙� + �)� (Eq. 2.3)

That is, the response consists of a decaying (��∙��∙�) sine with a phase shift ().

Note that in the examples in this section, the denominators of the fractions that give X(s) all have 1 for

the coefficient of the highest order derivative. Often it is the case that this coefficient is not 1. For

example, for the mck-system discussed at the first of this chapter, with a step force input (F0·u(t)) and

zero initial conditions

�(�) =��

� ∙�� + � ∙� + �

The coefficient of s2 is m. To make the above analysis apply, one can simply multiply the numerator and

denominator of the above fraction by 1/m. Thus

�(�) =�� ⁄

�� + (� �⁄ )� + � �⁄

Now the results of the analysis of the second-order system above can be directly applied to this mck-

system, with care being taken to use F0/m for F0 in the analysis and c/m for a and k/m for b. This type of

manipulation of fractions is very common in controls, so you should review the algebra of fractions and

be very adept at their manipulation, as illustrated in the foregoing analysis.

The case of complex roots is called an underdamped second order.

Transfer function

Control systems make use of what is called the transfer function of a system. As you have seen in

Chapter 1, an important graphical tool in controls is the block diagram. Each block, with the exception

of a summing block, takes one input, operates on it, and produces an output. Thus the block transfers

the input to the output. So the contents of a block are called the transfer function of that component.

As an example, with the mck-system, the applied, external force produces a displacement. So the

cause—the input—is the force and the effect—the displacement—is the output. The transfer function is

simply the quotient of effect/cause, output/input. The transfer function is usually given the symbol G.

Thus we might write


2-9

��(�) =�(�)

�(�)=

1

� ∙�� + � ∙� + �=

1 �⁄

�� + (� �⁄ ) ∙� + � �⁄=

1/�

(�/�) ∙�� + (�/�) ∙� + 1

All of these variants are useful for one purpose or another.

Final Value Theorem

As you have seen from the underdamped second order analysis, finding the inverse Laplace transform to

solve an ODE can be a tedious process. Often one is not interested in the time history of a system’s

motion but simply in its final value, where the system winds up after the dynamics have damped out.

There is a useful theorem for Laplace transforms that is given as the last entry of Table 2.2. This

theorem, the Final Value Theorem, allows a user to answer the question of where a system winds up

without calculating the inverse Laplace transform of a system. Take the mck-system above as an

example. Subject it to a unit step input, so F(s) = 1/s. Then

�(�) =1

� ∙(� ∙�� + � ∙� + �)

Using the FVT from Table 2.2

�(∞) = lim�→�

� ∙�(�) = lim�→�

�1

� ∙�� + � ∙� + �� =

1

�

Problems

2.1 A first-order system has the ODE

40 ∙∆ℎ̇ + ∆ℎ = 0

The ODE is written in the variable Δh, so the “motion” of the system is the time history ∆ℎ =

∆ℎ(�). Since the right-hand side of this equation is 0, F(t) = 0, i.e. there is no “force” being applied

to the system. Work out the system response to an initial non-zero ∆ℎ(0) = 1. Plot the time

history using Matlab. About how long does it take the system to get to equilibrium?

(We will encounter this system later. It represents a tank-level system, and h is the level of a

liquid in the tank. There is a reason that the system is written in the deviation of the tank level,

Δh, rather than in the tank level itself, h, as we shall see in the following Chapter.)

2.2 Repeat Problem 2.1, but start with 0 initial conditions. Apply a “force” of 4·u(t) to the system.

Plot the system response. About how long does it take the system to get to equilibrium?

2.3 Use Laplace transforms to calculate the unit step response of the system whose ODE is

2 ∙�̈ + 120 ∙�̇ + 80 ∙� = �(�)

The initial conditions for the system are all 0. Plot the response using Matlab.



2-10

2 ∙�̈ + 16 ∙�̇ + 32 ∙� = �(�)



2 ∙�̈ + �̇ + 16 ∙� = �(�)



2 ∙�̈ + 16 ∙� = �(�)


Documents

Control Systems Engineering - Cal Polyfowen/me422/BookUpThroughChap2.pdfControl Systems Engineering A Practical Approach by Frank Owen, PhD, P.E. Mechanical Engineering Department